Bond dissociation enthalpy of $E - H$ ( $E =$ element) bonds is given below. Which of the compounds will act as strongest reducing agent?
$\begin{matrix} C o m p o u n d & = & N H_{3} & P H_{3} & A s H_{3} & S b H_{3} Δ_{d i s s {(E - H)}^{H} (k J {m o l}^{- 1})} & = & 389 & 322 & 297 & 255 \end{matrix}$

N H_{3}

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P H_{3}

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A s H_{3}

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S b H_{3}

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Solution

The correct option is D $S b H_{3}$
As $E - H$ bond dissociation energy of $S b H_{3}$ is smallest among given compound so it can easily release H so we can say that it will be act like good reducing agent Weaker the $E - H$ bond, stronger the reducing agent i.e., $S b H_{3}$ .

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Similar questions

Compound	$N H_{3}$	$P H_{3}$	$A s H_{3}$	$S b H_{3}$
$Δ_{d i s s}$ $(E - H) / k J m o l^{- 1}$	389	322	297	255

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

1. S b H_{3} > N H_{3} > A s H_{3} > P H_{3}

2. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

3. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

4. N H_{3} > P H_{3} > A s H_{3} > S b H_{3}

{N H}_{3}, {P H}_{3}, A s H_{3}

S b H_{3}

N H_{3}, P H_{3}, A s H_{3}, S b H_{3}

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